Question

Hello, can you please help me solve question number 9

Answer 1

Given:

The equation is,

`2\cos ^2x+9\sin x=3\sin ^2x`

Step-by-step explanation:

Simplify the equation by using trigonometric identity.

`\begin{gathered} 2(1-\sin ^2x)+9\sin x=3\sin ^2x \\ 2-2\sin ^2x+9\sin x=3\sin ^2x \\ 5\sin ^2-9\sin x-2=0 \end{gathered}`

Assume sin x = t, then

`5t^2-9t-2=0`

Solve the equation by splitting the middle term.

`\begin{gathered} 5t^2-10t+t-2=0 \\ 5t(t-2)+1(t-2)=0 \\ (5t+1)(t-2)=0 \\ t=-(1)/(5),2 \end{gathered}`

So,

`\sin x=-(1)/(5)\text{ or sin x = 2}`

There is no possible value of x, for sin x = 2.

Determine the value of x by using sin x = -1/5.

`\begin{gathered} \sin x=-(1)/(5) \\ x=\sin ^(-1)(-(1)/(5)) \\ \approx-0.2014,-2.9402 \end{gathered}`

So possible values of x are -0.2014 and -2.9402.